This question came to mind when I was looking at this (http://www.stat.purdue.edu/~dasgupta/publications/tr02-03.pdf) paper by Professor Anirban DasGupta. In the last section, a couple of specific examples of his 'unified' method to find the sums of infinite series is pressented. In equation (34), he states that the following series:

$ 1/(1\cdot2) + 1/(3\cdot4) + 1/(5\cdot6) + ... 1/(2n\cdot(2n-1)) = log(2) $ (Note that $\{n\to\infty}$ again). I was wondering If it's possible to find the sum if the values of the denominators of the terms are squared.

The second sum which you have written does not sum to $\log 2$. It is an alternating series which sums to 1.
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Peter LuthyMay 11 '10 at 16:59

Maple gives the value $-3+\pi^2/3$. More generally, let $f(r) = 1/2^r + 1/12^r + 1/30^r + \cdots$, so your sum is $f(2)$. Then Maple gives $f(3) = 10-\pi^2, f(4) = -35+10\pi^2/3+\pi^4/45$, and in general $f(r)$ is a linear combination with rational coefficients of even powers of $\pi$. There appears to be some general theory at work here but I don't know what it is, which is why this is a comment and not an answer.
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Michael LugoMay 11 '10 at 16:59

1

Michael, your "general theory" is simply that 1/(x(x+1))^n is a sum of a bunch of x^j and (x+1)^j, for j varying over even integers between -n and -1. Just subtract off the principal parts of the Laurent series around 0 and -1; that gives an entire function which decays at infinity, hence equals zero.
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David HansenMay 11 '10 at 17:09

1

That being said, my original comment is in error; I was summing every term, and we only want every other term.
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Michael LugoMay 11 '10 at 17:39

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Ah, I see, you have written the general term down incorrectly, Max. It is not $1/(n(n+1))^2$ . It should be $1/(2n(2n-1))^2$.
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Peter LuthyMay 11 '10 at 17:49

haha I had just written this up a second before the site indicated you posted your solution... here is what I wrote: Write $\frac{1}{n^2(n+1)^2}=\left(\frac{1}{n}-\frac{1}{n+1}\right)^2=\frac{1}{n^2}-\fr‌​ac{2}{n(n+1)}+\frac{1}{(n+1)^2}$. Summing the first term gives $\pi^2/6$, the last term gives $-1+\pi^2/6$, and the middle term is an alternating seriesgives $-2$, so that the whole sum is $\frac{\pi^2}{3}-3$.
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Peter LuthyMay 11 '10 at 16:57

Gerald, Max wrote the incorrect general term, hence my error. $\displaystyle{\sum_{n=1}^\infty\frac{1}{((2n-1)2n)^2}=\sum_{n=1}^\infty\frac{1}‌​{(2n-1)^2}-2\sum_{n=1}^\infty\frac{1}{((2n-1)2n)^2}+\sum_{n=1}^\infty\frac{1}{(2n‌​)^2}}$. The first and second sums add to give $\displaystyle{\sum_{n=1}^\infty\frac{1}{n^2}}$, which then gives Michael Greenblatt's result.
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Peter LuthyMay 11 '10 at 17:56